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Balance the following equations by ion electron method. (i) KMnO_(4) + SnCl_(2) + HCl to MnCl_(2) + SnCl_(4) + H_(2)O + KCl (ii)C_(2)O_(4)^(2-) + Cr_(2)O_(7)^(2-) to Cr^(3+) + CO_(2) (in acidic medium) (iii) Na_(2)S_(2)O_(3) + I_(2) to Na_(2)S_(4)O_(6) + NaI Zn + NO_(3)^(-) to Zn^(2+) + NO(in acidic medium) |
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Answer» Solution :Oxidation half reaction: (loss of electrons) `overset(+2)(SnCl_(2)) to overset(+4)(SnCl_(4)) + 2e^(-)`(i) Reduction half reaction: (gain of electrons) `overset(+7)(KMnO_(4)) + 5E^(-) to overset(+2)(MnCl_(2))`(2) Add `H_(2)O` to balance hydrogen atoms. `KMnO_(4) + 5e^(-) to overset(+2)(MnCl_(2)) + 4H_(2)O`......(3) Add HCl to balance hydrogen atoms `KMnO_(4) + 5e^(-) + 8HCl to MnCl_(2) + 4H_(2)O`........(4) To equalize the number of ELECTRON equation `(1)xx5` and equation `(2)xx2` `5SnCl_(2) to 5SnCl_(2) + cancel(10e^(-))` `2KMnO_(4) + 16HCl + cancel(10e^(-)) to 2MnCl_(2) + 4H_(2)O + 2KCl` `2KMnO_(4) + 5SnCl_(2) + 16HCl to 5SnCl_(4) + 2MnCl_(4) + 2MnCl_(2) + 4H_(2)O + 2KCl` (ii)`C_(2)O_(4)^(2-) + Cr_(2)O_(7)^(2-) to Cr^(+3) + CO_(2)` (in acidic medium) Oxidation half reaction: `UNDERSET(+3)(C_(2)O_(4)^(2-)) to underset(+4)(2CO_(2)) + 2e^(-)` ......(1) Reduction half reaction: `underset(+6)(Cr_(2)O_(7)^(2-)) + 6e^(-) to 2Cr^(+3)`.....(2) To balance oxygen atoms, `H_(2)O` is added on RHS of equation (2) `Cr_(2)O_(7)^(2-) + 6e^(-) to 2Cr^(+3) + 7H_(2)O`......(3) To balance Hydrogen atom, `H^(+)` is added on LHS of equation (1) `C_(2)O_(4)^(2-) + 14H^(+) to 2CO_(2) + 2e^(-)`.....(4) To equalize the number of electrons gained and lost, multiply the equation `(4) xx 3`. (4) `=gt 3C_(2)O_(4)^(2-) + 14H^(+) to 6CO_(2) + cancel(6e^(-))` `Cr_(2)O_(7)^(2-) + cancel(6e^(-)) to 2Cr^(3+) + 7H_(2)O` `Cr_(2)O_(7)^(2-) + 3C_(2)O_(4)^(2-) 14H^(+) to 2Cr^(3+) + 6CO_(2) + 7H_(2)O` (iii) `Na_(2)S_(2)O_(3) + I_(2) to Na_(2)S_(4)O_(6) + NaI`(in acid medium) Oxidation half reaction: (Loss of electron) `Na_(2)S_(2)O_(3) to Na_(2)S_(4)O_(6) + 2e^(-)`......(i) Reduction half reaction: (Gain of electron) `I_(2) + 2e^(-) to 2NaI`.....(2) Adding (1) and (2) `Na_(2)S_(2)O_(3) to Na_(2)S_(4)O_(6) + cancel(2e^(-))` `I_(2) + cancel(2e^(-)) to NaI` `Na_(2)S_(2)O_(3) + I_(2) to Na_(2)S_(4)O_(6) + 2NaI` To balance oxygen, `2Na_(2)S_(2)O_(3) + I_(2) to Na_(2)S_(4)O_(6) + 2NaI` In acidic medium (iv)`Zn + NO_(3)^(-) to Zn^(2+) + NO` Half reaction medium `overset(o)(Z)n to Zn^(2+)`.....(1) `overset(+5)(N)O_(3)^(-) to overset(+2)(N)O`.....(2) `(1)=gt Zn to Zn^(2+) + 2e^(-)`.....(3) `(2)=gtNO_(3)^(-) + 4H^(+) to NO + 2H_(2)O`.....(4) `(3)xx3=gt3Zn to 3Zn^(2+) cancel(6e^(-))`....(5) `(4)xx2=gt 2NO_(3)^(-) + cancel(6e^(-)) + 8H^(+) to 2NO + 4H_(2)O`.....(6) `3Zn + 2NO_(3)^(-) + 8H^(+) to 3Zn^(2+) + 2NO + 4H_(2)O` |
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